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walters

  • 2 years ago

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  1. walters
    • 2 years ago
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    jjj let the mapping \[\tau _{ab}\] for a,b element R ,maps the reals by the ruel \[\tau _{ab}:x \rightarrow ax+b.Let G={\tau _{ab}:a \neq 0}\] Determine whether or not G is an abelian group under the composition of mappings

  2. walters
    • 2 years ago
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    oops ignore jjj

  3. KingGeorge
    • 2 years ago
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    Well, let \(\tau_{ab},\tau_{cd}\in G\). Then \[\tau_{ab}\circ\tau_{cd}(x)=\tau_{ab}(cx+d)=acx+ad+b.\]Also, \[\tau_{cd}\circ\tau_{ab}(x)=\tau_{cd}(ax+b)=cax+bc+d.\]Since \(acx+ad+b\neq acx+bc+d\), it must be that \(G\) is not abelian.

  4. walters
    • 2 years ago
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    am i not suppose to check whether is a group using the condition before i can conclude that is an abelian group

  5. KingGeorge
    • 2 years ago
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    It just asked if it was an abelian group. Since the elements don't commute, it can't be an abelian group even if it is a group. However, if we want to check it's a group, we can do that as well. Have you shown any of the axioms for proving it's a group yet?

  6. walters
    • 2 years ago
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    i am failing to show the axioms

  7. KingGeorge
    • 2 years ago
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    First, we can see that \(G\) is closed under composition. \[\tau_{ab}\circ\tau_{cd}(x)=\tau_{ab}(cx+d)=acx+ad+b=\tau_{(ac)(ad+b)}.\]We also have the identity \(\tau_{10}\). Now we just have to check associativity and inverses.

  8. walters
    • 2 years ago
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    why do u chose to use |dw:1360875075213:dw|

  9. KingGeorge
    • 2 years ago
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    "\(\circ\)" is a relatively common notation to denote the composition of two functions. You can also use \(\tau_{ab}(\tau_{cd}(x))\) is you prefer.

  10. walters
    • 2 years ago
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    can u also use * if u wnt

  11. KingGeorge
    • 2 years ago
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    For inverses, we want \[\tau_{ab}\circ\tau_{cd}(x)=acx+ad+b=x\]for some \(c,d\in\mathbb{R}\). We can see that \(c=a^{-1}\) and \(d=a^{-1}(-b)\) work for this direction. For the other, \[\tau_{cd}\circ\tau_{ab}(x)=cax+bc+d=a^{-1}ax+ba^{-1}-a^{-1}b=x,\]so we do indeed have inverses.

  12. KingGeorge
    • 2 years ago
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    If you make it clear your group function is composition of functions, you can use * as well.

  13. KingGeorge
    • 2 years ago
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    As for associativity, I'm just going to use the fact that composition of functions is always associative. If you have not heard of that yet, we can still go over how to prove this particular function is associative.

  14. walters
    • 2 years ago
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    please do

  15. KingGeorge
    • 2 years ago
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    let \(\tau_{ab},\tau_{cd},\tau_{ef}\in G\). Then\[\tau_{ab}\circ(\tau_{cd}\circ\tau_{ef})(x)=\tau_{ab}=\tau_{ab}(cex+cf+d)=acex+acf+ad+b\]and\[(\tau_{ab}\circ\tau_{cd})\circ\tau_{ef}(x)=(\tau_{(ac)(ad+b)})(ex+f)=acex+acf+ad+b\]Since these are equal, it's associative, and we have a group.

  16. walters
    • 2 years ago
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    k it is a group but not an abelian group

  17. KingGeorge
    • 2 years ago
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    Yup.

  18. walters
    • 2 years ago
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    thnx

  19. KingGeorge
    • 2 years ago
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    You're welcome.

  20. walters
    • 2 years ago
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    can we verify identity

  21. KingGeorge
    • 2 years ago
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    \[\tau_{10}(x)=1x+0=x\]

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